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I read that one can prove by AM-GM-inequality that for all $a,b,c\in\mathbb{R}_+$ we have that $$11(a^6 + b^6 + c^6) + 40abc(ab^2 + bc^2 + ca^2) \ge 51abc(a^2b + b^2c + c^2a)$$ How this can be done? Is it possible to prove stronger inequalities like $$10(a^6 + b^6 + c^6) + 41abc(ab^2 + bc^2 + ca^2) \ge 51abc(a^2b + b^2c + c^2a),$$ or even $$a^6 + b^6 + c^6 + 50abc(ab^2 + bc^2 + ca^2) \ge 51abc(a^2b + b^2c + c^2a)$$ without computer?

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The basic idea is that one can apply (weighted) AM-GM to take convex combinations of exponent vectors. This technique is also called "bunching," or Muirhead's inequality: en.wikipedia.org/wiki/Muirhead's_inequality –  Qiaochu Yuan Nov 21 '10 at 12:46
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up vote 2 down vote accepted

You want to apply AM-GM to some of the terms on the LHS to get some of the terms on the RHS. A naive way to do this such as

$$b^6 + c^6 + 2a^2 b^3 c \ge 4 a b^3 c^2$$

uses too many of the $11$ terms relative to the $40$ terms and only proves the weaker inequality

$$a^6 + b^6 + c^6 + abc(ab^2 + bc^2 + ca^2) \ge 2abc(a^2 b + b^2 c + c^2 a)$$

after cyclic summation. A slightly better combination of terms is

$$5a^6 + c^6 + 12a^2 b^3 c \ge 18 a^3 b^2 c$$

but this still uses too many of the $11$ terms relative to the $40$ terms and proves the inequality

$$a^6 + b^6 + c^6 + 2abc(ab^2 + bc^2 + ca^2) \ge 3abc(a^2 b + b^2 c + c^2 a)$$

after cyclic summation. You get a strong enough result by mixing some of the $40$ terms with each other to get

$$4a^6 + 12a^2 b^3 c + 4a^3 b c^2 \ge 20 a^3 b^2 c$$

which proves

$$a^6 + b^6 + c^6 + 4abc(ab^2 + bc^2 + ca^2) \ge 5abc(a^2 b + b^2 c + c^2 a).$$

Now multiply by $10$ and apply AM-GM to the remaining terms to conclude. The question of whether one can do better than this using AM-GM is a geometric question about using convex linear combinations of the vectors $(6, 0, 0), (0, 6, 0), (0, 0, 6), (2, 3, 1), (3, 1, 2), (1, 2, 3)$ to obtain the vectors $(3, 2, 1), (2, 1, 3), (1, 3, 2)$ and one might be able to prove that the above result is optimal.

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I don't have enough karma to comment, so I'll just drop this here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=154225 discusses the problem.

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